第四次

来源：互联网发布：农村淘宝发展前景分析编辑：程序博客网时间：2024/06/10 06:14

http://acm.csu.edu.cn/OnlineJudge/problem.php?cid=2101&pid=6

题意：
狗追小偷，如果找到了他的足迹，则速度变为原来2倍，有若干个出口，问最后小偷能否顺利逃脱。
思路：
枚举每一个出口，从小偷开始bfs，如果下一个点满足条件（小偷到这个点的距离+到出口距离<狗到这个点距离+到出口距离+1 /2）则可以转移，如果能转移到出口则可以逃脱~

#include <stdio.h>int grid_width, grid_height;char grid_sym[2510];int grid_dist_kennel[2500];int grid_dist_thief[2500];int grid_dist_exit[2500];int exits[200];#define FFQC 2510int ffq_pop_idx = 0, ffq_push_idx = 0;int ffqp[FFQC], ffqd[FFQC];void ffq_push(int pos, int depth) {    ffqp[ffq_push_idx] = pos;    ffqd[ffq_push_idx] = depth;    ffq_push_idx = (ffq_push_idx + 1) % FFQC;}int ffq_pop(int *pos, int *depth) {    if (ffq_pop_idx == ffq_push_idx) return 0;    *pos = ffqp[ffq_pop_idx];    *depth = ffqd[ffq_pop_idx];    ffq_pop_idx = (ffq_pop_idx + 1) % FFQC;    return 1;}void flood_fill(int start_pos, int *distances, int(*should_visit)(int, int)) {    for (int i = 0; i < 2500; ++i) {        distances[i] = -1;    }    ffq_push(start_pos, 0);    int pos, depth;    while (ffq_pop(&pos, &depth)) {        if (distances[pos] == -1 && grid_sym[pos] != 'X' && should_visit(pos, depth)) {            distances[pos] = depth;            const int row = pos / grid_width;            const int col = pos % grid_width;            if (row > 0)               ffq_push(pos - grid_width, depth + 1);            if (col > 0)               ffq_push(pos - 1,          depth + 1);            if (col + 1 < grid_width)  ffq_push(pos + 1,          depth + 1);            if (row + 1 < grid_height) ffq_push(pos + grid_width, depth + 1);        }    }}int should_visit_thief(int pos, int depth) {    const int thief_exit_time = depth + grid_dist_exit[pos];    const int dog_exit_time = grid_dist_kennel[pos] + (grid_dist_exit[pos] + 1) / 2;    return thief_exit_time < dog_exit_time;}int should_visit_other(int pos, int depth) {    return 1;}void handle_case() {    int thief, kennel, num_exits = 0;    for (int i = grid_width * grid_height; i > 0; --i) {        switch (grid_sym[i]) {            case 'T': thief = i; break;            case 'K': kennel = i; break;            case 'E': exits[num_exits++] = i;        }    }    flood_fill(kennel, grid_dist_kennel, should_visit_other);    for (int i = 0; i < num_exits; ++i) {        flood_fill(exits[i], grid_dist_exit, should_visit_other);        flood_fill(thief, grid_dist_thief, should_visit_thief);        if (grid_dist_thief[exits[i]] >= 0) {            puts("KEEP IT");            return;        }    }    puts("DROP IT");}int main() {    while (scanf("%d %d ", &grid_width, &grid_height) == 2 && grid_width >= 3 && grid_height >= 3) {        for (int row = 0; row < grid_height; ++row) {            fgets(&grid_sym[row * grid_width], 60, stdin);        }        handle_case();    }}

http://acm.csu.edu.cn/OnlineJudge/problem.php?cid=2101&pid=0
题意：一棵满二叉树代表一个家谱，定义几种成员关系，任意给定2个序号和b的性别，问a和b的关系。
找到最近公共祖先，根据a 到最近公共祖先的距离和a与b的距离之差即可判断出到底是什么关系，最近公共祖先只要找规律去求即可~

#include <stdio.h>char *nth[] = {"1st", "2nd", "3rd"};char *removed[] = {"once", "twice", "thrice"};#define MF(m, f) sex == 'M' ? m : fvoid handle_case(int id_a, int id_b, char sex) {    int dist_a = 0, dist_b = 0;    while (id_a != id_b) {        if (id_a > id_b) dist_a++, id_a = (id_a - 1) / 2;        else             dist_b++, id_b = (id_b - 1) / 2;    }    int smaller = dist_a < dist_b ? dist_a : dist_b;    int delta = dist_a - smaller + dist_b - smaller;    if (dist_a == 0 && dist_b == 0) {        puts("self");    } else if (dist_a == 1 && dist_b == 1) {        puts(MF("brother", "sister"));    } else if (smaller <= 1 && delta <= 4) {        switch (delta) {            case 4: fputs("great-", stdout);            case 3: fputs("great-", stdout);            case 2: fputs("grand",  stdout);        }        if      (dist_a == 0) puts(MF("son",    "daughter"));        else if (dist_b == 0) puts(MF("father", "mother"));        else if (dist_a == 1) puts(MF("nephew", "niece"));        else                  puts(MF("uncle",  "aunt"));    } else if (smaller >= 2 && smaller <= 4 && delta <= 3) {        if (delta == 0) printf("%s cousin\n", nth[smaller - 2]);        else printf("%s cousin %s removed\n", nth[smaller - 2], removed[delta - 1]);    } else {        puts("kin");    }}int main(int argc, char** argv) {    int id_a, id_b;    char sex;    while (scanf("%d %d %c", &id_a, &id_b, &sex) && id_a >= 0) {        handle_case(id_a, id_b, sex);    }    return 0;}

B:二分答案

#include <math.h>#include <stdio.h>int main() {    double pi = acos(-1);    double precision = 0.00001 * pi / 180;    double d, h, x, n1, n2;    while (scanf("%lf %lf %lf %lf %lf", &d, &h, &x, &n1, &n2) && d) {        double phi_min = 0;        double phi_max = pi / 2;        while (phi_max - phi_min > precision) {            double phi = (phi_min + phi_max) / 2;            double theta1 = pi / 2 - phi;            double theta2 = asin(n2 * sin(theta1) / n1);            if (d * tan(theta1) + h * tan(theta2) > x) {                phi_min = phi;            } else {                phi_max = phi;            }        }        printf("%0.02lf\n", phi_min * 180 / pi);    }}

暴力，注意字典序的排序方法以及字符串从1开始的递归终止条件是>长度！！！！！

#include <bits/stdc++.h>using namespace std;char d[250][30],s[10][10];int w,dd;char m[10][10];int dir[8][2]={1,0,-1,0,0,1,0,-1,1,1,1,-1,-1,1,-1,-1};struct Geng{    char s[30];}ans[250];bool cmp(Geng g1,Geng g2){    int l = 0;    for(int i = 0 ; g1.s[i] != '\0' && g2.s[i] != '\0' ; i++) {        if(g1.s[i] != g2.s[i]) return g1.s[i] < g2.s[i];        l = i;    }    int l1, l2;    l1 = l2 = l + 1;    if(g1.s[l + 1] != '\0') l1++;    if(g2.s[l + 1] != '\0') l2++;    return l1 < l2;}int find(char s[10][10],char *d,int x,int y,int pos){    if(pos>strlen(d+1)){        return 1;    }    char now=s[x][y];    s[x][y]='0';    for(int i=0;i<8;i++){        int nx=x+dir[i][0];        int ny=y+dir[i][1];        if(s[nx][ny]=='q'){            if(strlen(d+pos)>=2&&d[pos]=='q'&&d[pos+1]=='u'){              if(find(s,d,nx,ny,pos+2)){                  return 1;              }              }        }            else if(s[nx][ny]==d[pos]){                if(find(s,d,nx,ny,pos+1)){                    return 1;                }            }    }    s[x][y]=now;    return 0;}int main(){    scanf("%d",&w);        for(int i=1;i<=w;i++){            scanf("%s",d[i]+1);        }    while(scanf("%d",&dd)!=EOF){        if(dd==0) break;        for(int i=0;i<=8;i++){            for(int j=0;j<=8;j++){                s[i][j]='0';            }        }        for(int i=1;i<=dd;i++){            scanf("%s",s[i]+1);        }        for(int i=1;i<=dd;i++){            for(int j=1;j<=strlen(s[i]+1);j++){                m[i][j]=s[i][j];            }        }        int kk=0;        for(int i=1;i<=w;i++){            int flag=0;            for(int j=1;j<=dd;j++){                for(int k=1;k<=dd;k++){                    if(s[j][k]=='q'){                       if(strlen(d[i]+1)>=2&&d[i][1]=='q'&&d[i][2]=='u'){                        for(int i1=1;i1<=dd;i1++){                        for(int j1=1;j1<=strlen(s[i1]+1);j1++){                            m[i1][j1]=s[i1][j1];                            }                        }                         if(find(m,d[i],j,k,3)){                            strcpy(ans[kk++].s,d[i]+1);flag=1;break;                         }                       }                    }                       else if(s[j][k]==d[i][1]){                        for(int i1=1;i1<=dd;i1++){                        for(int j1=1;j1<=strlen(s[i1]+1);j1++){                        m[i1][j1]=s[i1][j1];                            }                                }                        if(find(m,d[i],j,k,2)){                            strcpy(ans[kk++].s,d[i]+1);flag=1;break;                        }                    }                }                if(flag) break;            }        }//      for(int i = 0 ; i < kk ; i++) printf("%s\n", ans[i].s);//      rp        sort(ans,ans+kk,cmp);        for(int i=0;i<kk;i++){            printf("%s\n",ans[i].s);        }        puts("-");    }}/*1rprquit5rpritahqlnietepzrysgogwey0*//**************************************************************    Problem: 1719    User: ALPC_t_1    Language: C++    Result: Accepted    Time:0 ms    Memory:1508 kb****************************************************************/

没想到这题用string写还是蛮方便的，直接暴力查询就好了，不用像char那样先匹配到一起，然后再整体比对，substr函数第一个是首指针，第二个是长度，可以直接取出来这个substr 判断是不是与字典相等即可~~

#include <bits/stdc++.h>using namespace std;const int MAXN = 300;const int MAXM = 1000 + 5;int m;string allot[MAXM];map<char, string> ma;char str1[MAXM];string str2;void init(){    ma.clear();    for(int i = 0 ; i < 26 ; i++) {        char u = i + 'A';        scanf("%s", str1);        cin>>str2;        ma[u] = str2;    }    scanf("%d", &m);    for(int i = 0 ; i < m ; i++) cin >> allot[i];}string data[MAXM];string out[MAXM];string match(string u){    for(int i = 0 ; i < m ; i++) {        int ok = 1;        int now = 0;        for(int j = 0 ; j < allot[i].size() ; j++) {            int len = ma[allot[i][j]].size();            if(u.substr(now, len) != ma[allot[i][j]]) {                ok = 0;                break;            }            else now += len;            if(now > u.size()) {                ok = 0;                break;            }        }        if(now != u.size()) ok = 0;        if(ok == 1) return allot[i];    }    return "-1";}int main(){    init();    int n;      while(scanf("%d", &n) != EOF && n) {        int flag = 1;        int pos;        for(int i = 0 ; i < n ; i++) {            cin >> data[i];            out[i] = match(data[i]);            if(out[i] == "-1") {                if(flag == 1) {                pos = i;                flag = 0;                }            }        }        if(flag == 0) {            cout << data[pos] << " not in dictionary.\n";            continue;        }        for(int i = 0 ; i < n ; i++) {            if(i) cout << " ";            cout << out[i] ;        }        printf("\n");    }    return 0;}

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